Question

If 2 and 1 are the roots of ax^2+bx+2=0 find a and b

Answer 1

Taking 2 as the root of the given equation,


So,

=> ax^2 + bx + 2 = 0
=> a( 2 )^2 + b( 2 ) + 2 = 0
=> 4a + 2b + 2 = 0
$= > 4a + 2b = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - (1)$





Then, Taking 1 as root,


=> ax^2 + bx + 2 = 0
=> a( 1 )^2 + b( 1 ) + 2 = 0
=> a + b + 2 = 0
=> a = -2 - b



Putting the value of a in ( 1 )

4( - 2 - b ) + 2b = -2
=> -8 - 4b + 2b = -2
=> -2b = 6
=> b = -3





Now, a = - 2 - b
a = - 2 - ( - 3 )
a = 1


$\boxed{ \bold{ a = - 1 \: \: \: \: \: \: and \: \: \: \: \: \: \: \: b = -3 }}$

Answer 2

solution

to find the constant values a & b

ax²+bx+2=0

where the root of the equations are 2 and 1

there for at x=2

⇒ a(2)²+b(2)+2=0

a(4)+2b+2=0

4a+2b+2=0 equation ..........1

at x=1

⇒a(1)²+b(1)+2=0

a+b+2=0 equation ............2

by making a or b be the subject formula from 1 or 2

⇒a=-b-2 equation .......3

then we put 3 into 1

⇒4(-b-2)+2b=-2

-4b-8+2b=-2

-2b=-2+8

-2b=6

by dividing booth site by -2

⇒b=-3

then we put the value of b into eqn 3

⇒a=-2-b

a=-2-(-3)

a=-2+3

⇒a=1

hence the values of a & b are required as

a=1

b=-3